Open Sets Built from Smaller Pieces
Open sets can be large or complicated, but their membership is governed locally: around each of their points, they contain suitable open pieces. A basis formalizes this idea. It is a collection of open sets that serves as a supply of building blocks: every open set can be assembled as a union of members of the collection. The pieces may overlap, and an open set may have many different such representations.
Throughout, \((X,\mathcal{T})\) is a topological space. A topology \(\mathcal{T}\) is a collection of subsets of \(X\) that contains \(\varnothing\) and \(X\), is closed under arbitrary unions, and is closed under finite intersections. Its members are the open sets. We allow the union of an empty family, which is \(\varnothing\).
Because \(X\) itself is open, the definition implies that the basis elements cover \(X\): every point belongs to at least one member of \(\mathcal{B}\). When \(X=\varnothing\), the empty collection is a basis: its union is \(\varnothing=X\), and that is the only open set. A basis is tied to the topology it describes; the same collection need not be a basis for a different topology on the same underlying set.
The Union Description and Its Local Form
Proof. Every set in the displayed union is contained in \(U\), so the union is contained in \(U\). For the reverse inclusion, let \(x\in U\). Since \(U\) is open and \(\mathcal{B}\) is a basis, \(U\) is a union of members of \(\mathcal{B}\). Thus at least one basis element in that union contains \(x\); call it \(B\). Since the union equals \(U\), this \(B\) is contained in \(U\). Therefore \(x\) belongs to the union of all basis elements contained in \(U\). This proves the reverse inclusion and hence equality. If \(U=\varnothing\), no point needs to be considered, and the union on the right is empty. \(\square\)
This formula gives a canonical way to write \(U\) as a union: use every basis element that lies inside it. It also avoids a hidden choice issue. To show that every point of \(U\) lies in some suitable basis element, we do not need to select one such element simultaneously for each point; the formula takes the union of the entire subcollection of suitable elements.
Proof. Suppose first that \(A\) is open. By the definition of a basis, \(A\) is a union of basis elements. For each \(x\in A\), membership in that union means that \(x\) belongs to one of its basis elements \(B\), and that element is contained in \(A\). Conversely, suppose that for every \(x\in A\) such a basis element exists. Let \(\mathcal{C}=\{B\in\mathcal{B}:B\subseteq A\}\). Every member of \(\mathcal{C}\) is contained in \(A\), so \(\bigcup_{B\in\mathcal{C}}B\subseteq A\). The assumed local property says that every point of \(A\) belongs to some member of \(\mathcal{C}\), so \(A\subseteq\bigcup_{B\in\mathcal{C}}B\). Therefore \(A=\bigcup_{B\in\mathcal{C}}B\). Each member of \(\mathcal{C}\) is open, and arbitrary unions of open sets are open. Hence \(A\) is open. If \(A=\varnothing\), the condition is vacuous and \(A\) is open as required. \(\square\)
The local form says that a basis element can be found inside an open set around each of its points. The size and shape of the available pieces depend on the topology and the basis. In the real line the pieces can be intervals; in a discrete space they can be singletons; in an indiscrete space there may be only one nonempty piece available.
Intervals as a Basis for the Usual Real Line
Give \(\mathbb{R}\) its standard topology. Its open sets have the following local description: a set \(U\subseteq\mathbb{R}\) is open if for every \(x\in U\), some \(\varepsilon>0\) satisfies \((x-\varepsilon,x+\varepsilon)\subseteq U\).
Worked Example: Open Intervals on the Real Line
Consider the collection \(\mathcal{B}=\{(a,b):a,b\in\mathbb{R},\ a<b\}\). We verify that it is a basis for the standard topology.
Each interval \((a,b)\) is open in the standard topology. Indeed, if \(x\in(a,b)\), then both \(x-a\) and \(b-x\) are positive. Set \(\varepsilon=\min(x-a,b-x)/2\), which is positive. Then \(x-\varepsilon>a\) and \(x+\varepsilon<b\), so \((x-\varepsilon,x+\varepsilon)\subseteq(a,b)\).
Now let \(U\) be open in the standard topology. For each \(x\in U\), choose an \(\varepsilon_x>0\) such that \((x-\varepsilon_x,x+\varepsilon_x)\subseteq U\). This interval is a member of \(\mathcal{B}\) containing \(x\). Consequently, \(U\) is the union of the intervals from \(\mathcal{B}\) that are contained in \(U\), by the theorem above. In the other direction, any union of members of \(\mathcal{B}\) is open because each interval is open and arbitrary unions of open sets are open. Thus the intervals are a basis for the standard topology.
For a concrete open set, let \(U=(-1,1)\cup(3,4)\). It is a union of two basis elements, namely \((-1,1)\) and \((3,4)\). The local description works at particular points as well: \((-1/2,1/2)\) is a basis element containing \(0\) and contained in \(U\); \((13/4,15/4)\) is a basis element containing \(7/2\) and contained in \(U\). Both inclusions follow directly from their endpoints. The local form guarantees the same kind of interval exists at every other point of \(U\).
Check: the displayed intervals really give \(U\), since their union contains exactly the real numbers strictly between \(-1\) and \(1\), or strictly between \(3\) and \(4\). Neither endpoint is included, in agreement with the open-interval notation.
Singletons and the Discrete Topology
A basis can also consist of very small sets. The discrete topology on a set \(X\) is \(\mathcal{T}=\mathcal{P}(X)\), the collection of all subsets of \(X\). Every subset is open in this topology.
Worked Example: A Singleton Basis for a Finite Discrete Space
Let \(X=\{p,q,r\}\) with the discrete topology. Take \(\mathcal{B}=\{\{p\},\{q\},\{r\}\}\). Every member is open, since every subset of \(X\) is open.
For example, the subset \(A=\{p,r\}\) has the representation \(A=\{p\}\cup\{r\}\). The subset \(\{q\}\) is already a basis element, and the empty set is the union of the empty subcollection. More generally, for any \(A\subseteq X\), we have \(A=\bigcup_{x\in A}\{x\}\): each singleton in the union contains only a point of \(A\), and each point \(x\in A\) belongs to its singleton. Thus every open set is a union of basis elements, and \(\mathcal{B}\) is a basis.
Check: for \(A=\{p,r\}\), the union includes \(p\) and \(r\), and includes neither \(q\) nor any other point. The union is exactly \(A\), as required.
Here the local pieces separate points completely: at any point \(x\) of an open set \(A\), the singleton \(\{x\}\) is a basis element contained in \(A\). This is more local detail than the real-line interval basis usually provides, but both bases describe open sets by unions.
The Indiscrete Topology: One Basic Piece
At the opposite extreme, the indiscrete topology on a set \(X\) is \(\mathcal{T}=\{\varnothing,X\}\). When \(X\) is nonempty, the collection \(\mathcal{B}=\{X\}\) is a basis for this topology.
Worked Example: A Basis for the Indiscrete Topology
Let \(X=\{a,b,c\}\) with the indiscrete topology. The only open sets are \(\varnothing\) and \(X\). The open set \(\varnothing\) is the union of the empty subcollection of \(\mathcal{B}=\{X\}\), while \(X\) is the union of the subcollection \(\{X\}\). There are no other open sets to represent. Therefore \(\mathcal{B}\) is a basis.
Notice the difference from the discrete example. The only basis element containing \(b\) is \(X\), so there is no basis element containing \(b\) that is contained in the proper subset \(\{b\}\). This causes no problem: \(\{b\}\) is not open in the indiscrete topology. For every open set that does contain \(b\), namely \(X\), the basis element \(X\) is available.
Check: the two possible unions of members of \(\{X\}\) are the empty union \(\varnothing\) and the union \(X\). They are exactly the two open sets of the stated topology.
If \(X=\varnothing\), the indiscrete topology has just one open set, \(\varnothing=X\). The empty collection is then a basis, since its empty union is that open set. This handles the empty space without requiring a nonempty basis element.
Overlapping Pieces and a Common Pitfall
A basis is a collection of building blocks, not a partition of the space. A partition would require distinct pieces to be disjoint; a basis has no such requirement. Overlap is useful because it lets different local descriptions fit together around points in the same open set.
Common pitfall. One might incorrectly claim that the interval basis represents each open set in exactly one way, with disjoint basis elements. In the standard topology on \(\mathbb{R}\), the open set \((0,2)\) is itself a basis element, so one representation is \((0,2)\). A different representation is \((0,2)=(0,3/2)\cup(1/2,2)\). These two intervals overlap: their intersection is \((1/2,3/2)\), which is nonempty because, for example, \(1\) satisfies \(1/2<1<3/2\). Both intervals are contained in \((0,2)\), and their union covers every point strictly between \(0\) and \(2\). Thus the representation is valid but not disjoint or unique.
The useful test is not whether the basis pieces are disjoint. It is whether the topology’s open sets can be expressed as unions of the pieces. The local form makes that test precise for an already specified basis: at each point of an open set, some basis element fits inside the set. For the real line, those pieces are intervals; for the discrete topology, they can be singletons; for the indiscrete topology, the only nonempty piece may be the whole space.
Check Your Understanding
Use the union definition and the local form of a basis to answer these questions.
- In the standard topology on \(\mathbb{R}\), give an interval-basis element containing \(1\) and contained in \((0,2)\).
- For \(X=\{p,q,r\}\) with the discrete topology and singleton basis, write \(\{q,r\}\) as a union of basis elements.
- For a nonempty set \(X\) with the indiscrete topology, which open set is represented by the empty union, and which is represented by the basis element \(X\)?
- Explain why the local condition for \(A=\{b\}\) fails in the three-point indiscrete example, and why this is consistent with the proposition.
- Can a basis representation use overlapping sets? Give the two interval representations of \((0,2)\) from this tutorial and state their intersection.